# Logic Proof

Note by Alexander Sludds
4 years, 10 months ago

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The statement that you are trying to prove may be thought of as four conditionals (although two are redundant, due to contraposition): if J then J & L; if not J then not (J & L); if J & L then J; and if not (J & L) then not J. The second and third of these are obvious, because of the way & works. The first is also obvious, since if J then L (as given by line two), hence if J then J & L. The last one follows similarly: if not (J & L), then not J or not L (de Morgan), and if not L then not J (contrapositive of line two). Thus, not (J & L) always implies not J.

In a Fitch-style natural deduction proof, the only premise would be that J implies L. Proving the biconditional would simply involve two subproofs of the conditionals. The one showing that L & J implies J would be very short; the one for showing J implies J & L would also be quite short.

I don't know how it would be best tackled in Hilbert's axioms

- 4 years, 10 months ago

Note that $$(L \wedge J) \equiv J$$ implies $$J \Rightarrow (L \wedge J)$$, which implies $$J \Rightarrow L$$, which is one of the premises. You should be able to fill in the gaps from there.

- 4 years, 10 months ago

Your chain of "implies" is going in the wrong direction. You should not see what is implied by what you're trying to prove, or you might end up in a circular argument. (That is, unless they are bi-implies, or biconditionals, but as stated above they are not.)

- 4 years, 10 months ago

Careful. A circular argument would require treating the conclusion as a premise. This is not what was done! Note that the conclusion was manipulated in isolation. The point was to show (hint, really) that the second premise and the conclusion are logically equivalent; the first premise is irrelevant. One could also manipulate just the second premise in isolation and show the same thing.

- 4 years, 10 months ago